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The average of the three positive integers p, q, and r is 10. If it is given that p ≤ q ≤ r and the median of three numbers is p + 2, then what is the smallest possible value of r?
1. 22
2. 12
3. 8
4. 10

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Correct Answer - Option 2 : 12

Calculation:

Given: the average of the three positive integers p, q, and r is 10

So, p + q + r = 30             ....(1)
Given that p ≤ q ≤ r and the median of three numbers is p + 2

So, q = p + 2

 

Now, p + q + r = 30

⇒ p + p + 2 + r = 30

⇒ r = 28 - 2p


q ≤ r

⇒ p + 2 ≤ 28 - 2p

⇒ 3p ≤  26
⇒ p ≤ 8 (since it must be an integer)

Maximum value of p = 8

Maximum value of q = p + 2 = 8 + 2 = 10

Put the value of p and q in equation (1), we get

Minimum value of r = 30 - 10 - 8 = 12

So the maximum values of p & q are 8 and 10, making the minimum value of r is 12.

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